Question

A recent study reported that 73% of Americans could only converse in one language. A random sample of 130 Americans was randomly selected. What is the probability that 100 or fewer of these Americans could only converse in one language?

Answer 1

Answer:

Probability that 100 or fewer of these Americans could only converse in one language is 0.8599.

Step-by-step explanation:

We are given that a recent study reported that 73% of Americans could only converse in one language.

A random sample of 130 Americans was randomly selected.

Let $\hat p$ = sample proportion of Americans who could only converse in one language.

The z score probability distribution for sample proportion is given by;

                                 Z  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, p = population proportion of Americans who could only converse in one language = 73%

           $\hat p$ = sample proportion = $\frac{100}{130}$ = 0.77

           n = sample of Americans = 130

Now, probability that 100 or fewer of these Americans could only converse in one language is given by = P( $\hat p$ $\leq$ 0.77)

      P( $\hat p$ $\leq$ 0.77) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\leq$ $\frac{0.77-0.73}{\sqrt{\frac{0.77(1-0.77)}{130} } }$ ) = P(Z $\leq$ 1.08) = 0.8599                    

The above probability is calculated by looking at the value of x = 1.08 in the z table which has an area of 0.8599.